Heather R Posted February 19, 2015 Share Posted February 19, 2015 My fifth grader is in tears because her math program taught that the sum of angles in a triangle is always 180• without teaching *why* this is the case... Can someone help me with an age appropriate explanation? I am at my wit's end! (And even more ready to read the results of her psych ed testing...) Thanks, Heather Quote Link to comment Share on other sites More sharing options...
ikslo Posted February 19, 2015 Share Posted February 19, 2015 I would pull up a Khan Academy video such as this one for her to watch: http://www.khanacademy.org/math/algebra-basics/core-algebra-geometry/copy-of-triang_prop_tut/v/proof-sum-of-measures-of-angles-in-a-triangle-are-180 Quote Link to comment Share on other sites More sharing options...
Tsuga Posted February 19, 2015 Share Posted February 19, 2015 That would bring me to tears as well. It's so insulting when they ask you to think critically but when they can't explain, they just say "trust us". Hugs for your daughter. Quote Link to comment Share on other sites More sharing options...
justasque Posted February 19, 2015 Share Posted February 19, 2015 Take a piece of paper and cut out a triangle - any kind. Put a little dot in each angle. Then cut, starting in the middle of one side, to the middle of the triangle. Do this on each of the three sides. This will give you three pieces of paper, each one having one of the triangle's angles (with a dot, so you can easily know which ones are the original angles). Now, put together the three original angles, with sides touching and the points together - like putting three pieces of pie next to each other in a pan. You will see that the free edges make a straight line - that is, the angles add up to 180 degrees. Do it again with another shape of triangle. And again, and again. It should work every time. This does't TELL why, but it does SHOW why. Quote Link to comment Share on other sites More sharing options...
regentrude Posted February 19, 2015 Share Posted February 19, 2015 I am with your kid. Such relationships should be proved, not believed. Draw a triangle ABC. Draw a line through point A that is parallel to the side BC. For sake of explaining w/o picture, label two points on the line, one left of A "X", one right of A "Y". You see that the angle X-A-Y is 180, because that is a straight line. Angle XAY=180 is the sum of three angles: two angles the line makes with sides of the triangle, and the triangle angle at A XAY = angle XAB + angle BAC (the triangle angle at A) + angle YAC Knowing about angles on parallel lines intersected by a transversal, you can identify: XAB = ABC and YAC=ACB (because XY || BC) Thus ABC+BAC+ACB=180. This may sound complicated from reading but becomes very simple if you draw a picture and use colored markers. Quote Link to comment Share on other sites More sharing options...
Miss Tick Posted February 19, 2015 Share Posted February 19, 2015 Beast Academy uses a demonstration like justasque's, but it just has the student tear off all three angles and then snug them up to a ruler. We had a lot of unarmed triangles to clean up last week. Quote Link to comment Share on other sites More sharing options...
Heather R Posted February 19, 2015 Author Share Posted February 19, 2015 Thanks everyone! She is much happier now! Quote Link to comment Share on other sites More sharing options...
Lecka Posted February 19, 2015 Share Posted February 19, 2015 Is she in Geometry? This is something I learned in Geometry in 10th grade -- and our Geometry program did explain it (we tore triangles, and the teacher did proofs on the board also). Maybe you can look for a Geometry program to supplement? It would not have made sense to me, either. I needed to see proofs and the hands-on was nice, too. I think I was more needful of proofs, personally, though, that is when things clicked better for me. Quote Link to comment Share on other sites More sharing options...
kiana Posted February 19, 2015 Share Posted February 19, 2015 Interesting side note -- although on a flat surface the sum of the angles in a triangle is 180 degrees, this is not necessarily true on a curved surface. For example, if you draw a triangle on a sphere, with one point at the "north pole", and two points on the "equator" separated by 90 degrees, this triangle will have 3 right angles, and yet undeniably have 3 straight sides. Quote Link to comment Share on other sites More sharing options...
ErinE Posted February 19, 2015 Share Posted February 19, 2015 I am with your kid. Such relationships should be proved, not believed. Draw a triangle ABC. Draw a line through point A that is parallel to the side BC. For sake of explaining w/o picture, label two points on the line, one left of A "X", one right of A "Y". You see that the angle X-A-Y is 180, because that is a straight line. Angle XAY=180 is the sum of three angles: two angles the line makes with sides of the triangle, and the triangle angle at A XAY = angle XAB + angle BAC (the triangle angle at A) + angle YAC Knowing about angles on parallel lines intersected by a transversal, you can identify: XAB = ABC and YAC=ACB (because XY || BC) Thus ABC+BAC+ACB=180. This may sound complicated from reading but becomes very simple if you draw a picture and use colored markers. This is how AoPS Pre-Algebra proves it and it's a great explanation. Quote Link to comment Share on other sites More sharing options...
SparklyUnicorn Posted February 19, 2015 Share Posted February 19, 2015 Take a piece of paper and cut out a triangle - any kind. Put a little dot in each angle. Then cut, starting in the middle of one side, to the middle of the triangle. Do this on each of the three sides. This will give you three pieces of paper, each one having one of the triangle's angles (with a dot, so you can easily know which ones are the original angles). Now, put together the three original angles, with sides touching and the points together - like putting three pieces of pie next to each other in a pan. You will see that the free edges make a straight line - that is, the angles add up to 180 degrees. Do it again with another shape of triangle. And again, and again. It should work every time. This does't TELL why, but it does SHOW why. This! It's very cool! Quote Link to comment Share on other sites More sharing options...
Grover Posted February 19, 2015 Share Posted February 19, 2015 Interesting side note -- although on a flat surface the sum of the angles in a triangle is 180 degrees, this is not necessarily true on a curved surface. For example, if you draw a triangle on a sphere, with one point at the "north pole", and two points on the "equator" separated by 90 degrees, this triangle will have 3 right angles, and yet undeniably have 3 straight sides. I won a morobar from my maths teacher in high school with this information. Quote Link to comment Share on other sites More sharing options...
raptor_dad Posted February 20, 2015 Share Posted February 20, 2015 I am with your kid. Such relationships should be proved, not believed. Draw a triangle ABC. Draw a line through point A that is parallel to the side BC. For sake of explaining w/o picture, label two points on the line, one left of A "X", one right of A "Y". You see that the angle X-A-Y is 180, because that is a straight line. Angle XAY=180 is the sum of three angles: two angles the line makes with sides of the triangle, and the triangle angle at A XAY = angle XAB + angle BAC (the triangle angle at A) + angle YAC Knowing about angles on parallel lines intersected by a transversal, you can identify: XAB = ABC and YAC=ACB (because XY || BC) Thus ABC+BAC+ACB=180. This may sound complicated from reading but becomes very simple if you draw a picture and use colored markers. That is classic Euclid but seems rather heavy weight as an explanation for an elementary kid... why do I want to use these derived parallel line traversal facts... furthermore it doesn't anticipate any of the later facts I am interested in at the elementary level. I am a hacker not a mathematician so I would prove it rigorously using constructions and exterior angles. Ideally I would do this is LOGO, but it works as well with paper and pencil. Assume that we know a line is 180deg and a circle is 360deg. Draw a triangle ABC. Using a straight edge extend line AB as ABA', line BC as BCB', Line CA as CAC'. In order go in a circle the exterior angles must sum to 360deg. So <A'BC + <B'CA + <C'AB=360. All the interior and exterior angles of the triangle combine to form 3 lines so 3 *180. The interior angles alone are the combined angles minus the exterior angles or 3*180-360=180deg. Ergo the angles of a triangle sum to 180. The great thing is you can obviously generalize this for an arbitrary n-gon. N-sides*180 -360 or (N-2)*180. Even better as a VSL, you can easily visually prove any N-gon can be composed of N-2 triangles so if you know a triangle is 180deg the rest follows obviously. This requires slightly different assumptions and is much more visual but I think it is much clearer to elementary kids... or at least more VSLish kids. Quote Link to comment Share on other sites More sharing options...
Arcadia Posted February 20, 2015 Share Posted February 20, 2015 My older shows it by using proof 2 of the link (using parallel lines,alternate angles and angles on a straight line) http://www.cut-the-knot.org/triangle/pythpar/AnglesInTriangle.shtml Quote Link to comment Share on other sites More sharing options...
EndOfOrdinary Posted February 20, 2015 Share Posted February 20, 2015 Parallel lines were what made sense to my son as well. He is a big picture thinker thoug. My algebra self was much happier with the three angles approach. The three angles is only due to the transversals, but I hated Geometry so much that I could have cared less. Quote Link to comment Share on other sites More sharing options...
regentrude Posted February 20, 2015 Share Posted February 20, 2015 That is classic Euclid but seems rather heavy weight as an explanation for an elementary kid... why do I want to use these derived parallel line traversal facts... furthermore it doesn't anticipate any of the later facts I am interested in at the elementary level. I like your proof :-) But I don't think mine is "heavy weight" - it is the level at which that would be proved back home in 6th grade, so not that far off from the OP's 5th grader. Quote Link to comment Share on other sites More sharing options...
JDoe Posted February 20, 2015 Share Posted February 20, 2015 Why is the sum of angles in a triangle equal to 180•? Hmm with the danger of causing more "problems" I believe you can go on adding more corners, increasing by 180 for each additional corner. Thus Shape Sum of angles Triangle 180 Square 360 Pentagon 540 Hexagon 720 Heptagon 900 Octagon 1080 Nonagon 1260 Decagon 1440 .... Of course we are here looking at the inside angles only, but the sum of the outside angles will always be 360 + 180 x Number of corners. The difference between (sum of) inside and outside angles seem however to stay constant at exactly 720 degrees, no matter how many corners we add. :) Merely a conjecture on my part, but maybe some geometry buff can either prove it wrong, or provide proof. :) PS. Proof for the square (or more generally parallelogram) is of course in line with Regentrude's above, just way simpler. Coincidentally, it seems one could then move from that proof to proving the 180 of the triangle. Quote Link to comment Share on other sites More sharing options...
JDoe Posted February 21, 2015 Share Posted February 21, 2015 Elaborated PS, that any triangle can be constructed via parallelograms can be demonstrated by going the other way and showing how any triangle can be converted into a parallelogram by rotating around a baseline midpoint. Quote Link to comment Share on other sites More sharing options...
JDoe Posted February 21, 2015 Share Posted February 21, 2015 Seems like the above would prove the general case also of the n-gon. Addition of an extra corner will always add 180 to both inside and outside angles, which may be seen simplest by starting with the simples plane figure the bilateral. Combination of two lines AB and BA. (Sum of inside angles =0, Outside angles =720) Addition of point C instantly causes both inside and outside angles to increase by 180, consideration of the above, and fact that only angles next to newly introduced point can be affected proves the general case. As a plus, this could be illustrated using a rubber band and a pencil with Regentrude's proof. Quote Link to comment Share on other sites More sharing options...
kiwik Posted February 21, 2015 Share Posted February 21, 2015 Interesting. I was never told and never thought to ask. I would have used a triangle inside a circle if I had to develop a proof though. Quote Link to comment Share on other sites More sharing options...
JDoe Posted February 21, 2015 Share Posted February 21, 2015 In order go in a circle the exterior angles must sum to 360deg. So <A'BC + <B'CA + <C'AB=360. Put a small circle in each corner of your triangle, each circle will then be the sum of interior and exterior angles of that corner. That is sum of interior and exterior angles of each corner will be 360 degrees. Triangle has three corners (3x360) and we know sum of the interior angles are 180 (regentrude's proof or teacher say so), thus sum of exterior angles MUST be 900 degrees in any triangle. Quote Link to comment Share on other sites More sharing options...
regentrude Posted February 21, 2015 Share Posted February 21, 2015 Put a small circle in each corner of your triangle, each circle will then be the sum of interior and exterior angles of that corner. That is sum of interior and exterior angles of each corner will be 360 degrees. Triangle has three corners (3x360) and we know sum of the interior angles are 180 (regentrude's proof or teacher say so), thus sum of exterior angles MUST be 900 degrees in any triangle. Caution with terminology! I need to point out that your use of the term "exterior angle" is NOT what is typically defined as "exterior angle" of a polygon in geometry. You use the term for 360 degrees minus inside angle. Any definition I have seen refers to the angle that is created when one side of the polygon is extended beyond the vertex; exterior angle is the angle between that line and the adjacent side of the polygon. These are two completely different things. The exterior angles (with the common definition, not yours) add to 360 degrees for any polygon, irrespective of number of vertices. Quote Link to comment Share on other sites More sharing options...
JDoe Posted February 21, 2015 Share Posted February 21, 2015 Caution with terminology! I need to point out that your use of the term "exterior angle" is NOT what is typically defined as "exterior angle" of a polygon in geometry. You use the term for 360 degrees minus inside angle. Any definition I have seen refers to the angle that is created when one side of the polygon is extended beyond the vertex; exterior angle is the angle between that line and the adjacent side of the polygon. These are two completely different things. The exterior angles (with the common definition, not yours) add to 360 degrees for any polygon, irrespective of number of vertices. Thank you for pointing out my sad lack of a decent geometry education, and thus improving upon my little knowledge. To RaptorDad I extend my sincerest apologies for my mistake. Illustration of Standard definition as per Regentrude's comment My nonstandard definition would add 180 degrees to the external angle. A little more on the confusion: http://mathworld.wolfram.com/ExteriorAngle.html Quote Link to comment Share on other sites More sharing options...
Alessandra Posted February 21, 2015 Share Posted February 21, 2015 Nothing mathematical to add here, except to say that this sort of thread is one reason I love WTM. You guys are great! Quote Link to comment Share on other sites More sharing options...
Heather R Posted February 22, 2015 Author Share Posted February 22, 2015 Yes, I agree Alessandra! Such a smart and helpful bunch! Quote Link to comment Share on other sites More sharing options...
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